If {\sec ^2 30^{\circ}+\tan ^2 60^{\circ}}/{\cos ^2 45^{\circ}+ \sin ^2 90^{\circ}}=\frac{p}{q} and G.C.D. of p, q is 1 , then \frac{p^2+q^2}{p^2-q^2}=

TS ICET 26 May 2023 Shift 1 Solved Paper

Question : 123 of 200

Marks: +1, -0

\;{\sec ^2 30^{∘}+\tan ^2 60^{∘}}/{\cos ^2 45^{∘}+ \sin \;^2 90^{∘}}=\;{p}/{q}

and G.C.D. of

p, q

is 1 , then

\;{p^2+q^2}/{p^2-q^2}=

Go to Question:

Prev Question

Next Question